Connexions module: m39533 1 Mechanical Properties of Matter: deformation of materials Free High School Science Texts Project This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License 1 Introduction In this chapter we will look at some mechanical (physical) properties of various materials that we use. The mechanical properties of a material are those properties that are aected by forces being applied to the material. These properties are important to consider when we are constructing buildings, structures or modes of transport like an aeroplane. 2 Deformation of materials 2.1 Hooke's Law Deformation (change of shape) of a solid is caused by a force that can either be compressive or tensile when applied in one direction (plane). Compressive forces try to compress the object (make it smaller or more compact) while tensile forces try to tear it apart. We can study these eects by looking at what happens when you compress or expand a spring. Hooke's Law relates the restoring force of a spring to its displacement from equilibrium length. The equilibrium length of a spring is its length when no forces are applied to it. When a force is applied to a spring, e.g., by attaching a weight to one end, the spring will expand and become longer. The dierence between the new length and the equilibrium length is the displacement. note: Hooke's law is named after the seventeenth century physicist Robert Hooke who discovered it in 1660 (18 July 1635-3 March 1703). Denition 1: Hooke's Law In an elastic spring, the extension varies linearly with the force applied. where F is the restoring force in newtons (N), k is the spring constant in N m 1 and x is the displacement of the spring from its equilibrium length in metres (m). Version 1.1: Aug 3, 2011 6:00 am GMT-5 http://creativecommons.org/licenses/by/3.0/
Connexions module: m39533 2 Figure 1: Hooke's Law - the relationship between a spring's restoring force and its displacement from equilibrium length. 2.1.1 Experiment : Hooke's Law Aim: Verify Hooke's Law. Apparatus: weights spring ruler Method: 1. Set up a spring vertically in such a way that you are able to hang weights from it. 2. Measure the equilibrium length, x 0, of the spring (i.e. the length of the spring when nothing is attached to it). 3. Measure the extension of the spring for a range of dierent weights. Note: the extension is the dierence between the spring's equilibrium length and the new length when a weight is attached to it, x x 0. 4. Draw a table of force (weight) in newtons and corresponding extension. 5. Draw a graph of force versus extension for your experiment. Conclusions: 1. What do you observe about the relationship between the applied force and the extension? 2. Determine the gradient (slope) of the graph. 3. Now calculate the spring constant for your spring. Phet simulation for Hooke's Law This media object is a Flash object. Please view or download it at <mass-spring-lab.swf> Figure 2 Khan academy video on springs and Hooke's law This media object is a Flash object. Please view or download it at <http://www.youtube.com/v/zzwuhs9ldby&rel=0&hl=en_us&feature=player_embedded&version=3> Figure 3
Connexions module: m39533 3 Exercise 1: Hooke's Law I (Solution on p. 5.) A spring is extended by 7 cm by a force of 56 N. Calculate the spring constant for this spring. Exercise 2: Hooke's Law II (Solution on p. 5.) A spring of length 20cm stretches to 24cm when a load of 0,6N is applied to it. 1. Calculate the spring constant for the spring. 2. Determine the extension of the spring if a load of 0,5N is applied to it. Exercise 3: Hooke's Law III (Solution on p. 5.) A spring has a spring constant of 400 N.m 1. By how much will it stretch if a load of 50 N is applied to it? 2.2 Deviation from Hooke's Law We know that if you have a small spring and you pull it apart too much it stops 'working'. It bends out of shape and loses its springiness. When this happens, Hooke's Law no longer applies, the spring's behaviour deviates from Hooke's Law. Depending on what type of material we are dealing with, the manner in which it deviates from Hooke's Law is dierent. We give classify materials by this deviation. The following graphs show the relationship between force and extension for dierent materials and they all deviate from Hooke's Law. Remember that a straight line show proportionality so as soon as the graph is no longer a straight line, Hooke's Law no longer applies. 2.2.1 Brittle material Figure 4: A hard, brittle substance This graph shows the relationship between force and extension for a brittle, but strong material. Note that there is very little extension for a large force but then the material suddenly fractures. Brittleness is the property of a material that makes it break easily without bending. Have you ever dropped something made of glass and seen it shatter? Glass does this because it is brittle. 2.2.2 Plastic material Figure 5: A plastic material's response to an applied force.
Connexions module: m39533 4 Here the graph shows the relationship between force and extension for a plastic material. The material extends under a small force but it does not fracture easily, and it does not return to its original length when the force is removed. 2.2.3 Ductile material Figure 6: A ductile substance. In this graph the relationship between force and extension is for a material that is ductile. The material shows plastic behaviour over a range of forces before the material nally fractures. Ductility is the ability of a material to be stretched into a new shape without breaking. Ductility is one of the characteristic properties of metals. A good example of this is aluminium, many things are made of aluminium. Aluminium is used for making everything from cooldrink cans to aeroplane parts and even engine blocks for cars. Think about squashing and bending a cooldrink can. Brittleness is the opposite of ductility. When a material reaches a point where Hooke's Law is no longer valid, we say it has reached its limit of proportionality. After this point, the material will not return to its original shape after the force has been removed. We say it has reached its elastic limit. Denition 2: Elastic limit The elastic limit is the point beyond which permanent deformation takes place. Denition 3: Limit of proportionality The limit of proportionality is the point beyond which Hooke's Law is no longer obeyed. 2.2.3.1 Hooke's Law and deformation of materials 1. What causes deformation? 2. Describe Hooke's Law in words and mathematically. 3. List similarities and dierences between ductile, brittle and plastic (polymeric) materials, with specic reference to their force-extension graphs. 4. Describe what is meant by the elastic limit. 5. Describe what is meant by the limit of proportionality. 6. A spring of length 15 cm stretches to 27 cm when a load of 0,4 N is applied to it. a. Calculate the spring constant for the spring. b. Determine the extension of the spring if a load of 0,35 N is applied to it. 7. A spring has a spring constant of 200 N.m 1. By how much will it stretch if a load of 25 N is applied to it? 8. A spring of length 20 cm stretches to 24 cm when a load of 0,6 N is applied to it. a. Calculate the spring constant for the spring. b. Determine the extension of the spring if a load of 0,8 N is applied to it.
Connexions module: m39533 5 Solutions to Exercises in this Module Solution to Exercise 1 (p. 3) Step 1. 56 = k 0, 07 (1) Solution to Exercise 2 (p. 3) k = 56 0,07 = 800N m 1 (2) Step 1. We know: F = 0,6 N The equilibrium spring length is 20 cm The expanded spring length is 24 cm Step 2. First we need to calculate the displacement of the spring from its equilibrium length: x = 24cm 20cm = 4cm = 0, 04m Now use Hooke's Law to nd the spring constant: (3) Step 3. F = 0,5 N We know from the rst part of the question that k = -15 N.m 1 So, using Hooke's Law: 0, 6 = k 0, 04 (4) k = 15N.m 1 Solution to Exercise 3 (p. 3) Step 1. x = F k = 0,5 15 = 0, 033m = 3, 3cm 50 = ( 400) x x = 50 400 = 0, 125m = 12, 5cm (5) (6) (7)